Question

If $f\left(x\right)=(x-2{)}^2{x}^n,n\in N,$ then $f\left(x\right)$ has/have

• A. local maximum at $x=2\forall n\in N$
• B. local minimum at $x=2\forall n\in N,n>1$
• C. local maximum at $x=0$ if $n$ is even
• D. local minimum at $x=0$ if $n$ is even
  

Answer 1

Answer: (B), (D)

local minimum at $x=0$ if $n$ is even


$f\left(x\right)=(x-2{)}^2{x}^n$
Differentiate w.r.t. $x$
$f^{\prime }\left(x\right)=(x-2{)}^2\cdot n{x}^{n-1}+x^n\cdot 2(x-2)\Rightarrow f^{\prime }(x)=x^{n-1}(x-2\left)\right[(x-2)n+2x]\Rightarrow f^{\prime }(x)=x^{n-1}(x-2\left)\right[(2+n)x-2n]$
From $f^{\prime }\left(x\right)=0,$ for $n\in N,n>1$
We get
$x=0,2,\frac{2n}{n+2}$
$\frac{2n}{n+2}$ lies in between $0$ and $2$


From above sign change of $f^{\prime }\left(x\right)$
local minimum at $x=0$ if $n$ is even and local minimum at $x=2\forall n\in N,n>1$